How do you factor #6x^2-13x-28#?

For your #x-#terms you could have: #2x# & #3x# or #6x# & #x#

For your constants, you need two numbers that have a product of #-28#. This means one must be positive and one must be negative. Since the #x#-term = #-13x# is negative, the larger constant needs to be negative.
Here are your choices: #1# & #-28#, #2# & #-14#, #4# & #-7#

Since the #x#-term = #-13x# is fairly small, select the constants that are close to each other: #4# & #-7#

Form possible factors:
#(6x +4)(2x - 7)# gives the #x#-term #= -42x + 8x = -34x# too big

#(3x+4)(2x - 7)# gives the #x#-term #= -21x + 8x = -13x# correct

Therefore the factors are: #(3x+4)(2x-7)#

Given:

#6x^2-13x-28#

Use an AC method:

Find a pair of factors of #AC = 6*28 = 168# which differ by #B=13#.

[[ Note that we look for a given difference rather than sum, since the sign on the constant term is negative ]]

Since the difference we want is odd, one of these factors must contain all of the powers of #2# in #168# - that is be a multiple of #8#.

We find that the pair #21, 8# works.

Use this pair to split the middle term and factor by grouping:

#6x^2-13x-28 = 6x^2-21x+8x-28#

#color(white)(6x^2-13x-28) = (6x^2-21x)+(8x-28)#

#color(white)(6x^2-13x-28) = 3x(2x-7)+4(2x-7)#

#color(white)(6x^2-13x-28) = (3x+4)(2x-7)#